# Cauchy sequences

#### steven187

hello all

I found this rather interesting
suppose that a sequence $${x_{n}}$$ satisfies

$$|x_{n+1}-x_{n}|<\frac{1}{n+1}$$ $$\forall n\epsilon N$$

how couldnt the sequence $${x_{n}}$$ not be cauchy? I tried to think of some examples to disprove it but i didnt achieve anything doing that, please help

thanxs

#### HallsofIvy

Homework Helper
An obvious one: just add those differences. Let xn be $$\sum_{i=1}^n \frac{1}{n}$$. That series does not converge and so the sequence of partial sums is not Cauchy.

#### saltydog

Homework Helper
I'm hesitant to challenge you Hall but feel compelled to do so at grave risk as Analysis is not my strong suite. Alright nothing is but I digress.

It looks like a Cauchy sequence to me based on the definition:

The sequence: $\{x_n}\}$ is a Cauchy sequence if given $\epsilon$ there exists N such that for all m,n$\leq$N we have:

$$|x_m-x_n|<\epsilon$$

For the sequence:

$$|x_{n+1}-x_{n}|<\frac{1}{n+1}$$

I take $\epsilon=\frac{1}{N+1}$;

Thus for all n>N:

$$|x_{n+1}-x_n|<\frac{1}{n+1}<\frac{1}{N+1}=\epsilon$$

R

#### rachmaninoff

Thus for all n>N:
$$|x_{n+1}-x_n|<\frac{1}{n+1}<\frac{1}{N+1}=\epsilon$$
That's true; but look closely - is that really the same thing as

for all m,n$\geq$N we have:$$|x_m-x_n|<\epsilon$$
?

#### HallsofIvy

Homework Helper
For a given &epsilon;, you then choose a large enough N. No fair choosing N first and then finding &epsilon; so that N doesn't work!
You do remind me that just $$|x_{n+1}-x_n|< \epsilon$$ is not enough. You have to show that $$|x_n-x_m|< \epsilon$$ which is basically showing that the tail of the sum goes to 0.

#### saltydog

Homework Helper
rachmaninoff said:
That's true; but look closely - is that really the same thing as

?

I suppose not.

$$|x_{n+1}-x_n|<\frac{1}{n+1}$$

does not imply

$$|x_m-x_n|<\frac{1}{n+1}$$

Still a little unclear. I'll work on it. Thanks guys.

Seriously, why do I even bother with differential equations anyway . . .

R

#### rachmaninoff

Simply put: for given N, your

$$| x_m - x_n |$$,

$$\mbox{ as } m \rightarrow \infty$$,

is not guaranteed to be any better than the (divergent) harmonic series:

$$| x_m - x_n | < \sum_{k=n+1}^{\infty} \frac{1}{n} \rightarrow \infty$$

#### saltydog

Homework Helper
Alright, I'm interested in getting my math right.
Following Hall's example (a slight change):

Let:

$$x_n=\sum_{i=1}^{n+1} \frac{1}{i}$$

thus:

$$|x_{n+1}-x_n|=|\sum_{n=1}^{n+2} \frac{1}{i} - \sum_{n=1}^{n+1} \frac{1}{i}|=\frac{1}{n+2}$$

thus:

$$|x_{n+1}-x_n|=\frac{1}{n+2}<\frac{1}{n+1}$$

However, the partial sums of this harmonic series $\{x_n\}$ do not converge and thus cannot be a Cauchy sequence.

Now, I got a pot of spagetti to make . . .

#### steven187

hello all

nice stuff saltydog, i had a similar result, i think your fine with analysis, and hallsofivy your right it is an obvious one but the funny thing is i cant think of any other examples, do you know of any others?

steven

#### Kevin"12

hi guys i need bounded but not cauchy function can you help me asap?

#### alexfloo

Kevin, consider any bounded periodic function that is not constant (like, for instance, a sinusoid). It will never become and remain arbitrarily close to anything, but it does remain within fixed bounds.

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